Let $X$ be integrable and $\cal{A}, \cal{B} \subset \cal{F}$ sigma algebra. I want to show that if $\sigma(\cal{X}, \cal{A})$ is independent of $\cal{B}$, then the the conditional expectation $E[X\mid\sigma(\cal{A}, \cal{B})] $$= E[X\mid\cal{A}]$ a.s.
I think I have to show that some measures $\mu(F) = E[E[X\mid\mathcal{A}]\mid 1_F]$ and $\nu(F) = E[E[X\mid\sigma(\mathcal{A}, \mathcal{B})]\mid1_F]$ for $F \in\cal{F}$ agree on the $\pi$ system $\{A \cap B \mid A\in \mathcal{A} \text{ and }B \in \cal{B}\}$. Then the statement of the proof follows from properties of the conditional expectation. How is this proven?
Let $\mu (E)= \int_E XdP$ and $\nu (E)=\int_E E(X|\mathcal A) dP$. Apply $\pi -\lambda$ theorem. Note that $\int_{A\cap B} XdP =P(B) \int_A XdP$ and $\int_{A\cap B} E(X|\mathcal A) dP =P(B) \int_A E(X|\mathcal A) dP $ by hypothesis.